Pass the WGU Courses and Certificates Applied-Probability-and-Statistics Questions and answers with ExamsMirror

The standard error of the sample mean is calculated as SE = s/√n, where s is the sample standard deviation and n is the sample size. Since n appears in the denominator, increasing the sample size reduces the standard error, assuming the standard deviation stays constant. This reflects the fact that larger samples produce more stable estimates of the population mean. Option B is incorrect because decreasing sample size increases standard error. Option C is incorrect because a larger standard deviation increases standard error by increasing the numerator. Option D is incorrect because the mean affects the center of the distribution but does not directly determine the standard error. Standard error is about precision: smaller standard error means less sampling variability and a more precise estimate. That is why larger samples are preferred in inference. Study Guide references/topics: standard error, sample size, standard deviation, sampling variability.

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A probability or percentage describes how often an event is expected to occur in the long run. Here, 3% of homes are infected with termites. As a probability, this is 0.03, meaning about 3 out of every 100 homes would be expected to have termites. This is a small probability, so the event should be interpreted as unlikely. It is not impossible, because impossible would require a probability of 0%. It is also not likely, because likely events generally have probabilities substantially greater than 50%. Finally, it is not “as likely as unlikely,” because that language corresponds to a probability near 50%, where occurrence and nonoccurrence are approximately balanced. Since 3% is close to zero but not equal to zero, the technically correct interpretation is that termite infection is unlikely for a randomly selected home in the community. References/topics from the Study Guide: probability interpretation, percentages, unlikely events, complementary probability.

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Simpson’s paradox occurs when an association observed in aggregated data reverses or conflicts with the association observed within relevant subgroups. In this case, the overall recovery rate appears better for Treatment Y, since Y has a 90% overall recovery rate compared with 80% for Treatment X. However, after the data are divided by age group, Treatment X has a higher recovery rate in every age group. That reversal between the overall comparison and the subgroup comparisons is precisely the defining structure of Simpson’s paradox. Age group functions as the lurking or confounding variable because different age distributions across treatment groups can distort the overall recovery rates. Option B states a true overall fact, but it ignores the contradiction created by the age-specific rates. Option C is too general; simple differences between treatments do not necessarily create Simpson’s paradox. The key issue is inconsistency between aggregate and stratified results. References/topics from the Study Guide: two-variable data, confounding variables, conditional comparisons, Simpson’s paradox.

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A t-test is used when making inferences about a population mean and the population standard deviation, σ, is unknown. In practical settings, σ is rarely available, so the sample standard deviation, s, is used as an estimate. That substitution introduces additional uncertainty, which is why the t-distribution is used instead of the standard normal distribution. The t-distribution has heavier tails, especially for small samples, reflecting the extra variability caused by estimating σ from the sample. Option B describes a z-test setting, where the population standard deviation is known. Option C is incorrect because categorical population data are usually analyzed with proportions, chi-square tests, or related categorical procedures, not mean-based t-tests. Option D is invalid because t-tests have a clear inferential role. The controlling condition is unknown σ, with inference focused on means. Study Guide references/topics: t-tests, unknown population standard deviation, sample standard deviation, inferential statistics.

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The scatterplot shows a clear downward pattern: as x increases, y tends to decrease. This establishes a negative correlation. The strength of a correlation is judged by how closely the points follow a general linear pattern. Here, although the points are not perfectly aligned, they form a visible descending band from upper left to lower right. That pattern is much stronger than a weak association, where the points would appear widely scattered with only a slight trend. Therefore, the best description is a strong negative correlation. Options C and D are incorrect because they describe positive correlation, which would require y to increase as x increases. Option A correctly identifies the negative direction but understates the strength of the pattern. In correlation terminology, a strong negative value would have r substantially below 0 and closer to −1 than to 0. References/topics from the Study Guide: scatterplots, negative correlation, correlation strength, bivariate quantitative data.

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The equation y = 40x + 75 is written in slope-intercept form, y = mx + b. In this form, m represents the rate of change and b represents the initial value. The constant term 75 means the graph must cross the y-axis at 75, because the studio charges $75 before any hourly time is added. The coefficient 40 means that for every additional hour of studio time, the total cost increases by $40. Therefore, the correct graph must begin at y = 75 when x = 0 and rise at a steady linear rate of 40 dollars per hour. For example, at x = 6 hours, the model gives y = 40(6) + 75 = 315, so the line should pass near the point (6, 315). Among the displayed graphs, Graph 4 matches both the y-intercept and the slope. References/topics from the Study Guide: basic algebra, linear equations, slope-intercept form, interpreting rate of change.

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To multiply decimals, first multiply the numbers as whole numbers and then place the decimal point according to the total number of decimal places in the factors. The values are 0.6 and 0.25. Ignoring decimals temporarily, 6 × 25 = 150. The factor 0.6 has one decimal place, and 0.25 has two decimal places, for a total of three decimal places. Therefore, 150 becomes 0.150, which is equivalent to 0.15. A second way to verify the computation is to convert the decimals into fractions: 0.6 = 6/10 and 0.25 = 25/100. Multiplying gives (6 × 25)/(10 × 100) = 150/1000 = 0.15. This result is reasonable because one-fourth of 0.6 is 0.15. The other answer choices are too large for one-fourth of 0.6 or do not result from the correct decimal multiplication. References/topics from the Study Guide: decimal multiplication, fractions, equivalent forms, numerical operations.

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When two standard six-sided dice are rolled, there are 6 × 6 = 36 equally likely ordered outcomes. To get a sum of 2, the only possible ordered pair is (1, 1). No other pair of die faces can produce a total of 2 because 1 is the smallest value on each die. Therefore, there is 1 favorable outcome out of 36 total outcomes, giving a probability of 1/36. Option B, 1/6, is the probability of one specific result on a single die, not a two-dice sum. Option C, 2/36, would require two favorable ordered pairs, but only one exists. Option D, 1/12, is also too large because it corresponds to 3 outcomes out of 36. The key is to count ordered outcomes in the full two-dice sample space. Study Guide references/topics: dice probability, sample space, favorable outcomes, theoretical probability.

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A pie chart displays how a whole is divided into categorical parts, usually as percentages totaling 100%. The chart shows solar at 35%, hydropower at 30%, wind at 20%, geothermal at 10%, and biomass at 5%. The largest percentage is solar, so solar power contributes the most to total energy production. Option A is false because hydropower plus biomass equals 30% + 5% = 35%, which is equal to solar, not more than solar. Option B is false because geothermal is 10%, while biomass is only 5%; therefore, biomass contributes the least. Option C is false because biomass plus geothermal equals 5% + 10% = 15%, which is less than wind at 20%. The correct interpretation requires comparing category percentages accurately rather than relying on visual size alone. References/topics from the Study Guide: pie charts, categorical data, percentages, interpreting graphical displays.

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A chi-square test is commonly used to analyze categorical data, especially to determine whether two categorical variables are associated. In a chi-square test of independence, data are arranged in a contingency table, and observed cell counts are compared with expected cell counts under the assumption that the variables are independent. If the observed counts differ substantially from the expected counts, there is evidence of association between the categorical variables. For example, a chi-square test could examine whether voting preference is associated with age group or whether product preference differs by region. Option B describes comparing means, which is typically handled by a t-test or related procedure. Option C refers to regression, which models relationships between variables, often involving quantitative outcomes. Option D refers to variance, which is tested using different procedures depending on context. The key phrase is “categorical variables”: chi-square methods work with counts in categories. Study Guide references/topics: categorical data, contingency tables, chi-square test, association.

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